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(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\

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$(\frac{L_2}{L_1})^{2/3}$

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(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$,which implies $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma - 1}$.Since the container is cylindrical with cross-sectional area $A$,the volume $V$ can be expressed as $V = L 	imes A$,where $L$ is the length of the gas column.Therefore,$\frac{V_2}{V_1} = \frac{L_2 A}{L_1 A} = \frac{L_2}{L_1}$.For a monoatomic gas,the adiabatic index $\gamma = 5/3$.Substituting these values,we get $\frac{T_1}{T_2} = (\frac{L_2}{L_1})^{5/3 - 1} = (\frac{L_2}{L_1})^{2/3}$.

$A$ monoatomic gas is filled in a cylindrical container with a piston at temperature $T_1$. It undergoes an adiabatic expansion such that its temperature becomes $T_2$. If $L_1$ and $L_2$ are the lengths of the gas column before and after the expansion respectively,then $T_1/T_2$ is equal to:

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